Physics Ch 4&amp;amp;5 - Chapter 4 solved problems Physics...

Chapter 4 solved problems Physics 1301 Fall 2010, Professor Carlos R. Ordonez 6. Picture the Problem: The motion of the electron is depicted at right. Strategy:Use the given information to independently write the equations of motion in the xand ydirections. There will be a pair ofequations for the position of the particle (like equation 4-6), except theacceleration will not be the same as g.Use the equations of motion tofind the requested time and position information. Let the xdirectioncorrespond to horizontal, and the ydirection to vertical.Solution:1. (a)The horizontal motion is character- ized by constant velocity. Apply equation 4-6: ?096.20 0 cm2.95 10 s2.10 10 cm/sxxxtv2.(b)Use the time to find the vertical deflection , again using equation 4-6 except substituting for .ag 2100217292120 cm 0 cm5.30 10 cm/s(2.95 10 s)2.31 cmyyyyv ta tInsight:This problem is very much akin to projectile motion, with uniform acceleration in one direction and constant velocity the perpendicular direction. The only difference is the acceleration is 5.30×1017cm/s2upward instead of 9.81m/s2downward. 17. Picture the Problem: The trajectory of the climber is indicated inthe figure at right. Strategy:The 45direction of motion indicates that, just prior to landing, the climber is falling with a speed equal to his horizontal speed. Use this fact together with equation 4-7 (because the initial velocity is horizontal) find the height difference of the crevasse and the landing point of the climber. Solution:1. (a)Use the fact that 0yvvto find the time of flight: 27.8 m/s9.81 m/s0.80 syvtg  2.Find the vertical drop during the flight, which is also the height difference between the two sides of   222119.81 m/s0.80 s3.1 mh ygt0vax y

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the crevasse: 3. (b)Find the horizontal distance traveled:   7.8 m/s 0.80 s6.2 mxx v t4.The climber lands 6.2 – 2.8 m = 3.4 m beyond the far edge of the w= 2.8 m wide crevasse. Insight:The climber impacts the other side of the crevasse at about 25 mi/h (verify this for yourself!). It would bemuch safer to cross the crevasse with a ladder! 20. Picture the Problem: The pumpkin’s trajectory is depicted in the figure at right. Strategy: The horizontal velocity remains 3.3 m/s throughout the flight of the pumpkin, but the vertical velocity continually increases in the downward direction due to gravity. Use the given time information to find the vertical component of the velocity, and use it together with the horizontal component to find the direction and magnitude of the velocity vector. Equation 4-7 apply because theinitial velocity of the pumpkin is horizontal. Solution:1. (a)Determine the vertical component of the velocity:   29.81 m/s0.75 s7.4 m/syyvgtv2.Find the direction of the velocity vector: 17.4 m/stan66 or 66 below horizontal3.3 m/s 3.Find the magnitude of the velocity vector:

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